首页 > 代码库 > 2016 大连网赛---Weak Pair(dfs+树状数组)

2016 大连网赛---Weak Pair(dfs+树状数组)

2024-08-12 21:37:12   223人阅读

题目链接

http://acm.split.hdu.edu.cn/showproblem.php?pid=5877

 

Problem Description
You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a non-negative value ai is assigned.An ordered pair of nodes (u,v) is said to be weak if
  (1) u is an ancestor of v (Note: In this problem a node u is not considered an ancestor of itself);
  (2) au×av≤k.

Can you find the number of weak pairs in the tree?
 
Input
There are multiple cases in the data set.
  The first line of input contains an integer T denoting number of test cases.
  For each case, the first line contains two space-separated integers, N and k, respectively.
  The second line contains N space-separated integers, denoting a1 to aN.
  Each of the subsequent lines contains two space-separated integers defining an edge connecting nodes u and v , where node u is the parent of node v.

  Constrains: 
  1≤N≤105 
  0≤ai≤109 
  0≤k≤1018
 
Output
For each test case, print a single integer on a single line denoting the number of weak pairs in the tree.
 
Sample Input
1
2 3
1 2
1 2
 
Sample Output
1
 
题意：输入n，k  表示有n个节点的一棵树，然后输入n个节点的权值和n-1条边，求点对（u，v）的对数，满足：
      1、u是v的祖先节点。
      2、a[u]*a[v]<=k，a[]是存储权值的数组。
 
思路：从根节点开始向下深搜，每到一个点时计算sum+=Sum（a[i]）,Sum(x)表示大于等于x的个数，然后向树状数组中加入k/a[i]，继续递归深搜，退栈时从树状数组中减去k/a[i] ,这样可以保证树状数组中存的一直是一条到根节点的路径值。大题思路如上，这里要做一个离散化的处理，输入的权值<=1e9  k<=1e18  而只有1e5个点，所以可以离散到2*1e5 后处理；
 
题解中提示用treap计算大于等于x的个数，这样可以不需要进行离散化；
第一次自己做出深搜的题，挺高兴的^_^  看样子我对深搜有了一点认识了
 
代码如下：
#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <vector>#include <map>using namespace std;const long long maxn=200003;long long root;long long sum,k;long long in[100005];vector<long long>g[100005];long long a[100005];long long b[200005];long long c[200005];map<long long,long long>q;long long Lowbit(long long t){    return t&(t^(t-1));}void add(long long x,long long t){    while(x > 0)    {        c[x]+=t;        x -= Lowbit(x);    }}long long Sum(long long li){    long long s=0;    while(li<200005)    {        s+=c[li];        li=li+Lowbit(li);    }    return s;}void dfs(long long t){    long long n=g[t].size();    for(long long i=0;i<n;i++)    {        long long v=g[t][i];        sum+=(long long)Sum(q[a[v]]);        if(a[v]==0) add(maxn,1);        else   add(q[k/a[v]],1);        dfs(v);        if(a[v]==0) add(maxn,-1);        else   add(q[k/a[v]],-1);    }}int main(){    long long T,N;    scanf("%lld",&T);    while(T--)    {        q.clear();        memset(in,0,sizeof(in));        memset(c,0,sizeof(c));        memset(b,0,sizeof(b));        scanf("%lld%lld",&N,&k);        for(long long i=1;i<=N;i++)        {            scanf("%lld",&a[i]);            b[2*i-2]=a[i];            if(a[i]!=0)            b[2*i-1]=k/a[i];            g[i].clear();        }        sort(b,b+2*N);        long long tot=0,pre=-1;        for(long long i=0;i<2*N;i++)        {            if(b[i]!=pre)            {                pre=b[i];                q[pre]=++tot;            }        }        for(long long i=0;i<N-1;i++)        {            long long aa,bb;            scanf("%lld%lld",&aa,&bb);            g[aa].push_back(bb);            in[bb]++;        }        for(long long i=1;i<=N;i++)        if(in[i]==0) { root=i; break; }        sum=0;        if(a[root]==0) add(maxn,1);        else   add(q[k/a[root]],1);        dfs(root);        printf("%lld\n",sum);    }    return 0;}

 

 

2016 大连网赛---Weak Pair(dfs+树状数组)


联系
我们